Using l'hopitals rule, evaluate the following:
Lim x-->infinity
Inx/2(x)^1/2
Limx-->0 (1/x - 1/e^x -1)
I tried both multiple times and I'm still not getting an answer. For the first one I tried l'hopitals rule 3 times and I'm still getting infinity over infinity for an answer which isn't correct, and the second one is very confusing, please help!
To evaluate the limit using l'Hôpital's rule, we need to differentiate both the numerator and the denominator separately and see if the resulting limit is easier to evaluate.
Let's start with the first limit: Lim x-->infinity Inx/2(x)^1/2.
To apply l'Hôpital's rule, we need to differentiate the numerator and the denominator.
The derivative of Inx is 1/x (using the chain rule), and the derivative of 2(x)^1/2 is (1/2)(2)(x)^(-1/2) = (1/2)(x)^(-1/2).
Now, we have the limit of (1/x)/(1/2)(x)^(-1/2) as x approaches infinity. We can simplify this to (1/x)(2(x)^1/2) = 2(x)^(1/2)/(x) = 2(x)^(1/2-1).
As x approaches infinity, the exponent 1/2-1 becomes -1/2, so the limit simplifies to:
Lim x-->infinity 2(x)^(-1/2) = 2/((x)^1/2).
Now, as x approaches infinity, the denominator (x)^1/2 also approaches infinity, so the whole expression converges to zero.
Therefore, the answer to the first limit is 0.
Now, let's move on to the second limit: Lim x-->0 (1/x - 1/e^x -1).
To apply l'Hôpital's rule, we need to differentiate the numerator and the denominator.
The derivative of 1/x is -1/x^2, and the derivative of 1/e^x is (-e^x)(-1) = e^x.
Now, we have the limit of (-1/x^2)/(e^x) as x approaches 0.
Again, as x approaches 0, both the numerator and denominator approach 0, so we can apply l'Hôpital's rule again.
Differentiating the numerator and denominator one more time, we get the following:
The derivative of (-1/x^2) is (2/x^3), and the derivative of e^x is e^x.
Now, we have the limit of (2/x^3)/(e^x) as x approaches 0.
As x approaches 0, both the numerator and denominator approach 0, so we can apply l'Hopital's rule again.
Differentiating the numerator and denominator again, we get:
The derivative of (2/x^3) is (-6/x^4), and the derivative of e^x is e^x.
Now, we have the limit of (-6/x^4)/(e^x) as x approaches 0.
Again, as x approaches 0, both the numerator and denominator approach 0, so we can apply l'Hopital's rule one more time.
Differentiating the numerator and denominator once more, we get:
The derivative of (-6/x^4) is (24/x^5), and the derivative of e^x is e^x.
Now, we have the limit of (24/x^5)/(e^x) as x approaches 0.
Again, as x approaches 0, both the numerator and denominator approach 0, so we can now evaluate the limit without using l'Hopital's rule.
The limit becomes:
Lim x-->0 (24/x^5)/(e^x) = 24/(x^5 * e^x), and as x approaches 0, x^5 * e^x also approaches 0.
Therefore, the answer to the second limit is 0.
I hope this explanation helps you understand l'Hôpital's rule and how to apply it. Remember, if you're not getting the answer, check your algebraic manipulations and make sure you're applying the rule correctly.